Question

What is the pH of this solution?

Answer 1

pH = 11.216.

Step-by-step explanation:

Hello there!

In this case, according to the ionization of ammonia in aqueous solution:

`NH_3+H_2O\rightleftharpoons NH_4^++OH^-`

We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:

`Kb=([NH_4^+][OH^-])/([NH_3]) \\\\1.80x10^(-5)=(x*x)/(0.150-x)`

However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:

`1.80x10^(-5)=(x*x)/(0.150)\\\\x=\sqrt{1.80x10^(-5)*0.150}=1.643x10^(-3)M`

Which is also:

`[OH^-]=1.643x10^(-3)M`

Thereafter we can compute the pOH first:

`pOH=-log(1.643x10^(-3)M)\\\\pOH=2.784`

Finally, the pH turns out:

`pH=14-2.784\\\\pH=11.216`

Regards!